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The dimensions are $[m]=\mathsf{M}$, $[\alpha ]=\mathsf{M}{\mathsf{T}}^{-1}$, $[u_0]=\mathsf{L}{\mathsf{T}}^{-1}$, and $[g]=\mathsf{L}{\mathsf{T}}^{-2}$. Under the stated model, $m,\alpha ,u_0,g$ are the only parameters determining the ascent time. If $m^p{\alpha }^q{u}_0^r{g}^s$ is dimensionless, then $p+q=0$, $r+s=0$, and $-q-r-2s=0$, leaving one independent dimensionless variable. One convenient choice isThus dimensional analysis fixes the form, but Newton's second law must determine the function $f$.

Take upward velocity as positive. Throughout ascent, Newton's second law with linear drag gives $m\dot{v}=-mg-\alpha v$. An integrating factor or direct solution of this linear ordinary differential equation givesThe first zero of this velocity is the highest point, soThis vertical ascent under linear drag has $T<u_0/g$ for $u_0>0$, since $\log (1+\lambda )<\lambda$. The limit $\alpha \to 0$ recovers the no-drag ascent time $u_0/g$; $u_0=0$ gives $T=0$.

For constant downward gravitational acceleration $g$ and linear drag $-\alpha v$, the upward velocity solves $m\dot{v}=-mg-\alpha v$ with $v(0)=u_0\ge 0$. Its first zero occurs at the displayed time. The dimensionless control variable is $\alpha u_0/(mg)$, and the limit $\alpha \to 0$ gives $u_0/g$.